What is the solution for y^{''}-2y^'+2y=2x-2 ?

The solution for y^{''}-2y^'+2y=2x-2 is y=e^x(c_{1}cos(x)+c_{2}sin(x))+x

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y^{''}-2y^'+2y=2x-2

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x - 2

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + x

Solution steps

y^{''} (x) - 2 y^{'} (x) + 2 y = 2 x - 2

Solve linear ODE:

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + x

y = e^{x} (c_{1} cos (x) + c_{2} sin (x)) + x

x \to 0 + lim (2^{- \frac{1}{x}})

\int \frac{1}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}} d x

\int \frac{4 x ^{2} + x + 27}{x ^{3} + 9 x} d x

\frac{\partial}{\partial x} (2 x^{2} - 7 y^{9} - 4)

e^{x} sin (x) + e^{x} cos (x)

What is the solution for y^{''}-2y^'+2y=2x-2 ?
The solution for y^{''}-2y^'+2y=2x-2 is y=e^x(c_{1}cos(x)+c_{2}sin(x))+x